BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2, Problem 13RQ
To determine

Whether the statement, “if f is continuous at 5 and f(5)=2 and f(4)=3, then limx2f(4x211)=2” is true or false.

Expert Solution

Answer to Problem 13RQ

The statement is true.

Explanation of Solution

Theorem used: If f is continuous at b and limxag(x)=b, then limxaf(g(x))=f(b).

Calculation:

Given that, f is continuous at 5 and f(5)=2 and f(4)=3.

Here, f is a composition of two functions.

Take g(x)=4x211, a=2 and b=5.

Obtain the limit of the function g(x) as x approaches 2.

Here, g(x) is a polynomial. So it is continuous everywhere. That is, limx2g(x)=g(2).

limx2g(x)=g(2)=(4(2)211)=1611=5

Thus, limx2(4x211)=5.

Since f is continuous at 5 and limx2(4x211)=5 then by theorem stated above limx2f(4x211)=f(5) (1)

Substitute f(5)=2 in equation (1), limx2f(4x211)=2.

Hence, the required proof is obtained.

Note that, f(4)=3 is required to prove the result.

Thus, the given statement is true.

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